Degrees of freedom of Sample Variance of Residuals (Chi-Square distribution?

Question

In the context of jointly testing J linear restrictions, I am reviewing the distribution of the F-statistic, which is F(J, n-k). Below, R is a full row rank Jxk matrix, q is a Jx1 vector, and the null hypothesis tested is $H_0:R\beta-q=0$. We assume that residuals are homoskedastic and that they are conditionally normal.

I have that

$$F=(Rb-q)'[R(X'X)^{-1}R']^{-1}(Rb-q)/Js^2 \\=(Rb-q)'[R(X'X)^{-1}R']^{-1}(Rb-q)/J\sigma^2/(s^2/\sigma^2)$$ and this should be $\sim{F(J, n-k)}$

It should follow from the distributional result that, given two independent random scalars $x_1\sim{\chi^2(p_1)}$ and $x_2\sim{\chi^2(p_2)}$, it holds that

$$(x_1/p_1)/(x_2/p_2)\sim{F(p_1,p_2)}$$

In our case, we have that: $$x_1=(Rb-q)'{R(X'X)^{-1}R'(Rb-q)/\sigma^{2}}$$ $$p_1=J$$ $$x_2=s^2/\sigma^2$$ and then $p_2$ should be $n-k$. But here $x_2$ is not divided by $n-k$.

My intuition:

since we know that $(n-k)(s^2/\sigma^2)\sim{\chi^2(n-k)}$, it holds that $$s^2/\sigma^2\sim{\chi^2}(1)$$

In this way, $p_2$ would be just 1, when again it should be $n-k$. Can soebody shed some light on this?

Answer 1

Let $$ Q:= {(\mathbf R{\mathbf b }-\mathbf q)^\mathsf T\left[\mathbf R(\mathbf X^\mathsf T\mathbf X)^{-1}\mathbf R^\mathsf T\right]^{-1}(\mathbf R{\mathbf b }-\mathbf q)}.\tag 1$$

As outlined in this CV.SE post,

$$ \frac Q{\sigma^2}\sim{\chi^2}^\prime\left[r(\mathbf R) ,\frac{{(\mathbf R{\mathbf b }-\mathbf q)^\mathsf T\left[\mathbf R(\mathbf X^\mathsf T\mathbf X)^{-1}\mathbf R^\mathsf T\right]^{-1}(\mathbf R{\mathbf b }-\mathbf q)}}{2\sigma^2}\right].\tag 2$$

Also,

\begin{align}\textrm{SSE}/\sigma^2 &\sim {\chi^2}_{n-r(\mathbf X) },\tag{3.1}\\(n-r(\mathbf X) )s^2/\sigma^2 &\sim {\chi^2}_{n-r(\mathbf X)}.\tag{3.2}\end{align}

Reformulating $Q$ and $\textrm{SSE}$ as quadratics in $\mathbf y - \mathbf X \mathbf R^\mathsf T(\mathbf R\mathbf R^\mathsf T) ^{-1}\mathbf q,$ it can be shown that they are distributed independently (cf.$\rm [I],$ section $3.6,$ pp. $111-112$).

Now \begin{align}F &:= \frac{Q/r(\mathbf R) }{\textrm{SSE}/[n-r(\mathbf X) ]}\\ &= \frac{(Q/\sigma^2)/r(\mathbf R) }{(\textrm{SSE}/\sigma^2)/[n-r(\mathbf X) ]}\\&\sim \mathsf F^\prime\left[r(\mathbf R),n-r(\mathbf X), \frac{{(\mathbf R{\mathbf b }-\mathbf q)^\mathsf T\left[\mathbf R(\mathbf X^\mathsf T\mathbf X)^{-1}\mathbf R^\mathsf T\right]^{-1}(\mathbf R{\mathbf b }-\mathbf q)}}{2\sigma^2}\right].\tag 4\end{align}

Therefore, under the null hypothesis $\mathcal H_0: \mathbf R\boldsymbol\beta = \mathbf q, $

\begin{align}F &= \frac{Q}{r(\mathbf R) s^2}\\ &\sim \mathsf F_{r(\mathbf R), n-r(\mathbf X) }.\tag 5\end{align}

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Reference:

$\rm [I]$ Linear Models, S. R. Searle, John Wiley & Sons., $1971.$